Answer:
41 g
Explanation:
We have a buffer formed by a weak acid (C₆H₅COOH) and its conjugate base (C₆H₅COO⁻ coming from NaC₆H₅COO). We can find the concentration of C₆H₅COO⁻ (and therefore of NaC₆H₅COO) using the Henderson-Hasselbach equation.
pH = pKa + log [C₆H₅COO⁻]/[C₆H₅COOH]
pH - pKa = log [C₆H₅COO⁻] - log [C₆H₅COOH]
log [C₆H₅COO⁻] = pH - pKa + log [C₆H₅COOH]
log [C₆H₅COO⁻] = 3.87 - (-log 6.5 × 10⁻⁵) + log 0.40
[C₆H₅COO⁻] = [NaC₆H₅COO] = 0.19 M
We can find the mass of NaC₆H₅COO using the following expression.
M = mass NaC₆H₅COO / molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = M × molar mass NaC₆H₅COO × liters of solution
mass NaC₆H₅COO = 0.19 mol/L × 144.1032 g/mol × 1.5 L
mass NaC₆H₅COO = 41 g
Answer:
14 mL
Explanation:
To prepare a solution by a concentrated solution, we must use the equation:
C1xV1 = C2xV2, where <em>C</em> is the concentration, <em>V</em> is the volume, 1 is the initial solution and 2 the final solution.
The final solution must have 2 mL and a concentration of 350 pg/mL, and the initial solution has a concentration of 50 pg/mL.
Then:
50xV1 = 350x2
50xV1 = 700
V1 = 700/50
V1 = 14 mL
Answer:
Jack Beacuse...
Explanation:
the soil has water in it so it will be more than one matterial
Transform boundaries are places where plates slide sideways past each other. At transform boundaries lithosphere is neither created nor destroyed. Many transform boundaries are found on the sea floor, where they connect segments of diverging mid-ocean ridges. California's San Andreas fault is a transform boundary. So its true
